Find the gradient of $f(x, y, z) = -x + 3xy + 5xyz$ at $(-1, 2, 1)$. $\nabla f(-1, 2, 1) = ($
Solution: The gradient of a scalar field is all its partial derivatives put together into a vector. For a 3D scalar field, this looks like $\nabla f = (f_x, f_y, f_z)$. Let's find $f_x$, $f_y$, and $f_z$. $\begin{aligned} f_x &= \dfrac{\partial}{\partial x} \left[ -x + 3xy + 5xyz \right] \\ \\ &= -1 + 3y + 5yz \\ \\ f_y &= \dfrac{\partial}{\partial y} \left[ -x + 3xy + 5xyz \right] \\ \\ &= 3x + 5xz \\ \\ f_z &= \dfrac{\partial}{\partial z} \left[ -x + 3xy + 5xyz \right] \\ \\ &= 5xy \end{aligned}$ Now we can evaluate the partial derivatives we found at the point $(-1, 2, 1)$. $\begin{aligned} f_x(-1, 2, 1) &= -1 + 3y + 5yz \\ \\ &= -1 + 3(2) + 5(2)(1) = 15 \\ \\ f_y(-1, 2, 1) &= 3x + 5xz \\ \\ &= 3(-1) + 5(-1)(1) = -8 \\ \\ f_z(-1, 2, 1) &= 5xy \\ \\ &= 5(-1)(2) = -10 \end{aligned}$ The gradient of $f$ at $(-1, 2, 1)$ is $\nabla f = (15, -8, -10)$.